Chapter Four fluid flow mass, energy, Bernoulli and momentum

Chapter Four fluid flow mass, energy, Bernoulli and momentum

Qahtan A. Mahmood

4-1Conservation of Mass Principle

Consider a control volume of arbitrary shape, as shown in Fig (4-1).

Figure (4-1): the differential control volume and differential control volume

(Total mass entering CV)- (Total mass leaving CV) = Net change in mass within the CV

∫

Total mass within the CV:

∫

Rate of change of mass within the CV:

∫

Now consider mass flow into or out of the control volume through a differential area dA

on the control surface of a fixed control volume.

Let be the outward unit vector of dA normal to dA and be the flow velocity at dA

relative to a fixed coordinate system, as shown in Fig. (4-1). In general, the velocity may

cross dA at an angle .


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u

The mass flow rate through dA is proportional to the fluid density normal velocity u,

and the flow area dA, and can be expressed as,

Differential mass flow rate:

The net flow rate into or out of the control volume through the entire control surface is

obtained by integrating . Over the entire control surface,

Net mass flow rate:

∫

∫

∫

The conservation of mass relation for a fixed control volume can then be expressed as

∫

∫

The general conservation of mass relation can also be expressed as

∑∫

∑∫

∫

Or in mass flow rate


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∑

∑

∫

Example (4-1)

A 4ft high, 3ft diameter cylindrical water tank whose top is open to the atmosphere is

initially filled with water. Now the discharge plug near the bottom of the tank is pulled

out, and water jet whose diameter is 0.5 in streams out. The average velocity of the jet is

given by √ , where h is the height of water in the tank measured from the center

of the hole (a variable) and g is the gravitational acceleration. Determine how long it

will take for the water level in the tank to drop to 2 ft from the bottom.

Solution:

√

√

√

√

√


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∫

√

∫

√

√ √

√

√ √

√

4-2-Bernoulli equation The Bernoulli equation is an approximate relation between pressure, velocity, and

elevation, and is valid in regions of steady, incompressible low where net frictional

forces are negligible (4-2).

Figure (4-2) the force acting on a fluid particle along a streamline.

Applying Newton’s second law in the s-direction on a particle moving along a

streamline gives

∑

Two component of acceleration


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Substituting (4-12) in equation (4-11) yield

Canceling dA from each term and simplifying,

Noting

that and dividing each term by gives

By integration

∫

For incompressible

This is the famous Bernoulli equation, which is commonly used in fluid mechanics for

steady, incompressible flow.


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The Bernoulli equation can also be written between any two points on the same

streamline as

Static, Dynamic, and Stagnation Pressures

The kinetic and potential energies of the fluid can be converted to flow energy (and

vice versa) during flow, causing the pressure to change. This phenomenon can be made

more visible by multiplying the Bernoulli equation by the density

Each term in this equation has pressure units, and thus each term represents some kind

of pressure:

P is the static pressure; it is represents the actual thermodynamic pressure of the

fluid.

u22 is the dynamic pressure; it represents the pressure rise when the fluid in

motion is brought to a stop isentropically

gz is the hydrostatic pressure, which is not pressure in a real sense since its

value depends on the reference level selected; it accounts for the elevation effects,

i.e., of fluid weight on pressure

Limitations on the Use of the Bernoulli Equation

Steady flow: it should not be used during the transient start-up and shut-down

periods,


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Frictionless flow: (valve and sharp entrance are disturbs the streamlined structure

of flow)

No shaft work: pump, turbine, fan, or any other machine or impeller since such

devices destroy the streamlines

No heat transfer

Flow along a streamline: no irrotational region of the flow

Hydraulic Grade Line (HGL) and Energy Grade Line (EGL)

This is done by dividing each term of the Bernoulli equation by g to give

Where

P/g is the pressure head; it represents the height of a fluid column that produces

the static pressure P.

u2/2 g is the velocity head; it represents the elevation needed for a fluid to reach

the velocity u during frictionless free fall.

z is the elevation head; it represents the potential energy of the fluid

Also, H is the total head for the flow.

(4-22)


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Figure (4-4): HGL and EGL for frictionless flow in a duct

Example (4-2)

A large tank open to the atmosphere is filled with water to a height of 5 m from the

outlet tap. A tap near the bottom of the tank is now opened, and water flows out from

the smooth and rounded outlet. Determine the water velocity at the outlet.

Solution:

P1 = Patm (open to the atmosphere), u1 = 0 (the tank is large relative to the outlet) and

z2=0 Also, P2 = Patm (water discharges into the atmosphere).

Then the Bernoulli equation simplifies to

√ √

√

Example (4-3)


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A pressurized tank of water has a 10cmdiameter orifice at the bottom, where water

discharges to the atmosphere. The water level is 3 m above the outlet. The tank air

pressure above the water level is 300 kPa (absolute) while the atmospheric pressure is

100 kPa. Neglecting frictional effects, determine the initial discharge rate of water from

the tank.

Solution:

√ (

*

√ (

)

4-3-Conservation of Energy

One of the most fundamental laws in nature is the first law of thermodynamics, also

known as the conservation of energy principle.

The conservation of energy principle for any system can be expressed simply as

Then the conservation of energy for a fixed quantity of mass can be expressed in rate

form as (Fig. 4-5)


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Figure (4-5): A closed system

∫

Total energy consists of internal, kinetic and potential energies, and it is expressed on a

unit-mass basis as

Energy Transfer by Heat, Q

For adiabatic process Q=0

For isothermal process there is change in temperature

Energy Transfer by Work, W A system may involve numerous forms of work, and the total work can be expressed as

Shaft Work

Many flow systems involve a machine such as a pump, a turbine, a fan, or a compressor

whose shaft protrudes through the control surface, and the work transfer associated with

all such devices is simply referred to as shaft work Wshaft


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Where:

Work Done by Pressure Forces

Consider a gas being compressed in the piston-cylinder device shown in Fig. (4-6)

Where u= ds/dt is the piston velocity

So

Figure (4-6) piston-cylinder

Work done by pressure forces is positive when it is done on the system and negative

when it is done by the system,

∫

∫

Then the rate form of the conservation of energy relation for a closed system becomes


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(Net rate of energy into CV)+(time rate of change of CV)=net flow rate of energy CS

∫

∫

∫

∫

Substituting the surface integral for the rate of pressure work from Eq.(4-27) into Eq.

(4-30) and combining it with the surface integral on the right give

∫

∫

∫

Then equation (4-31) become

∫

∑

(

* ∑

∫

∑

(

) ∑

or

∫

∑

(

) ∑


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The last two equations are fairly general expressions of conservation of energy,

ENERGY ANALYSIS OF STEADY FLOWS

For steady flows, the time rate of change of the energy content of the control volume is

zero, and Eq. (4-34) simplifies to

∑

(

) ∑

Many practical problems involve just one inlet and one outlet. The mass flow rate for

such single-stream devices remains constant, and Eq. (4-34) reduces to

(

)

Divided by mass flowrate

(

)

and rearranging, the steady-flow energy

equation can also be expressed as


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For single-phase fluids (a gas or a liquid), we have

U2 - U1 = cv (T2 - T1)

Where cv is the constant-volume specific heat.

Noting that wshaft, net in = w shaft, in - w shaft, out = wpump - wturbine, the mechanical

energy balance can be written more explicitly as

Multiplying Eq. (4-39) by the mass flow rate m. gives

The total mechanical power loss which consists of pump, turbine losses and frictional

losses in the piping network. That is,

Divided equation (4-39) by g


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Where

Is the useful head delivered to the fluid by the pump. Because of irreversible losses in

the pump, , is less than

by the factor

. similarly

Is the extracted head removed from the fluid by the turbine. Because of irreversible

losses in the turbine, , e is greater than

by the factor

. Finally

Is the irreversible head loss between 1 and 2 due to all components of the piping system

other than the pump or turbine? Note that the head loss hL represents the frictional

losses associated with fluid flow in piping, and it does not include the losses that occur

within the pump or turbine due to the inefficiencies of these devices

Equation (4-41) is illustrated schematically in Fig. (4-7).


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Figure (4-7): Mechanical energy flow chart for a fluid flow system

Example (4-4)

The pump of a water distribution system is powered by a 15kW electric motor whose

efficiency is 90 percent (Fig. below). The water flow rate through the pump is 50 L/s.

The diameters of the inlet and outlet pipes are the same, and the elevation difference

across the pump is negligible. If the pressures at the inlet and outlet of the pump are

measured to be 100 kPa and 300 kPa (absolute), respectively, determine (a) the

mechanical efficiency of the pump and (b) the temperature rise of water as it flow

through the pump due to the mechanical inefficiency.

SOLUTION:


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u1=u2=0 and z1=z2 so

(

*

(

)(

*

Then the mechanical efficiency of the pump becomes

Of the 13.5kW mechanical power supplied by the pump, only 10 kW is imparted to the

fluid as mechanical energy. The remaining 3.5 kW is converted to thermal energy due to

frictional effects, and this “lost” mechanical energy manifests itself as a heating effect in

the fluid,

The temperature rise of water due to this mechanical inefficiency is determined from the

thermal energy balance,

Solving for T,

( )(

*

Example (4-5)


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In a hydroelectric power plant, 100 m3/s of water flow from an elevation of 120 m to a

turbine, where electric power is generated (Fig. below). The total irreversible head loss

in the piping system from point 1 to point 2 (excluding the turbine unit) is determined to

be 35 m. If the overall efficiency of the turbine–generator is 80 percent, estimate the

electric power output

Solution:

P1=P2=atm

u1=u2=0

Equation (4-41) reduce to

(

,

The electric power generated by the actual unit

4-4-Conservation of Momentum Principle Newton’s second law for a system of mass m subjected to a net force F is expressed as

∑

Where is the linear momentum of the system, Newton’s second law can be

expressed more generally as


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∑

∫

where is the mass of a differential volume element dV and is its

momentum.

(Sum of force acting CV)=(Rate of change of momentum of CV)+(net flow rate at CS)

∫

∫

∑

∫

∫

During steady flow, the amount of momentum within the control volume remains

constant (the second term of Eq. 4-45) is zero. It gives

∑ ∫

Mass flow rate across an inlet or outlet

∫

Momentum flow rate across a uniform inlet or outlet:

∫

so

∑ ∑

∑

Many practical problems involve just one inlet and one outlet, and Eq. (4-49) reduces to

∑


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Momentum Equation for Two and three dimensional flow along a streamlin

Consider the two dimensional system shown, since both momentum and force are vector

quantities, they can be resolving into components in the x and y directions

These components can be combined to give the

resultant force

√

And the angle of this force

Force exerted by a jet striking flat plate

Consider a jet striking a flat plate that may be perpendicular or inclined to the direction

of the jet.

The general term of the jet velocity component normal to the plate can be written as:

(

)


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The mass flow entering the control volume

If the plate is stationary:

Thus the rate of change of momentum normal to the plate

Force exerted normal to the plate = The rate of change of momentum normal to the

plate:

if the plate is stationary and inclined

if the plate is both stationary and perpendicular

Example (4-6)

A jet of water from a fixed nozzle has a diameter d of 25mm and strikes a flat plate at

angle of 30o to the normal to the plate. The velocity of the jet is 5m/s, and the surface

of the plate can be assumed to be frictionless. Calculate the force exerted normal to the

plate (a) if the plate is stationary, (b) if the plate is moving with velocity u of 2m/s in the

same direction as the jet.


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Solution:

a) Force exerted normal to the plate = The rate of change of momentum normal to the

plate:

if the plate is stationary and inclined

( )

(

)

if the plate is moving with velocity 2m/s

(

)

Example (4.7)

A reducing elbow is used to deflect water flow at a rate of 14 kg/s in a horizontal pipe

upward 30° while accelerating it (Fig. below). The elbow discharges water into the

atmosphere. The cross-sectional area of the elbow is 113 cm2 at the inlet and 7 cm

2 at

the outlet. The elevation difference between the centers of the outlet and the inlet is 30

cm. The weight of the elbow and the water in it is considered to be negligible.

Determine (a) the gage pressure at the center of the inlet of the elbow and (b) the

anchoring force needed to hold the elbow in place.


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level (z1 = 0) and noting that P2 = Patm, the Bernoulli equation for a streamline

going through the center of the elbow is expressed as

(b) The momentum equation for steady one-dimensional flow is

∑


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(

(

) +(

)

(

*

If we repeated example above for the figure below

Noting that the outlet velocity is negative since it is in the negative x-direction, we have

(

)

Chapter Four fluid flow mass, energy, Bernoulli and momentum

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Transcript of Chapter Four fluid flow mass, energy, Bernoulli and momentum